3.1.48 \(\int \frac {(e x)^m (a+b x^2)^p (A+B x^2)}{(c+d x^2)^2} \, dx\) [48]

Optimal. Leaf size=295 \[ \frac {(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^{1+p}}{2 c (b c-a d) e \left (c+d x^2\right )}-\frac {(a d (A d (1-m)+B c (1+m))-b c (A d (1-m-2 p)+B c (1+m+2 p))) (e x)^{1+m} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} F_1\left (\frac {1+m}{2};-p,1;\frac {3+m}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{2 c^2 d (b c-a d) e (1+m)}-\frac {b (B c-A d) (1+m+2 p) (e x)^{1+m} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\left (\frac {1+m}{2},-p;\frac {3+m}{2};-\frac {b x^2}{a}\right )}{2 c d (b c-a d) e (1+m)} \]

[Out]

1/2*(-A*d+B*c)*(e*x)^(1+m)*(b*x^2+a)^(1+p)/c/(-a*d+b*c)/e/(d*x^2+c)-1/2*(a*d*(A*d*(1-m)+B*c*(1+m))-b*c*(A*d*(1
-m-2*p)+B*c*(1+m+2*p)))*(e*x)^(1+m)*(b*x^2+a)^p*AppellF1(1/2+1/2*m,-p,1,3/2+1/2*m,-b*x^2/a,-d*x^2/c)/c^2/d/(-a
*d+b*c)/e/(1+m)/((1+b*x^2/a)^p)-1/2*b*(-A*d+B*c)*(1+m+2*p)*(e*x)^(1+m)*(b*x^2+a)^p*hypergeom([-p, 1/2+1/2*m],[
3/2+1/2*m],-b*x^2/a)/c/d/(-a*d+b*c)/e/(1+m)/((1+b*x^2/a)^p)

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Rubi [A]
time = 0.29, antiderivative size = 295, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {593, 598, 372, 371, 525, 524} \begin {gather*} -\frac {(e x)^{m+1} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} (a d (A d (1-m)+B c (m+1))-b c (A d (-m-2 p+1)+B c (m+2 p+1))) F_1\left (\frac {m+1}{2};-p,1;\frac {m+3}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{2 c^2 d e (m+1) (b c-a d)}-\frac {b (m+2 p+1) (e x)^{m+1} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} (B c-A d) \, _2F_1\left (\frac {m+1}{2},-p;\frac {m+3}{2};-\frac {b x^2}{a}\right )}{2 c d e (m+1) (b c-a d)}+\frac {(e x)^{m+1} \left (a+b x^2\right )^{p+1} (B c-A d)}{2 c e \left (c+d x^2\right ) (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((e*x)^m*(a + b*x^2)^p*(A + B*x^2))/(c + d*x^2)^2,x]

[Out]

((B*c - A*d)*(e*x)^(1 + m)*(a + b*x^2)^(1 + p))/(2*c*(b*c - a*d)*e*(c + d*x^2)) - ((a*d*(A*d*(1 - m) + B*c*(1
+ m)) - b*c*(A*d*(1 - m - 2*p) + B*c*(1 + m + 2*p)))*(e*x)^(1 + m)*(a + b*x^2)^p*AppellF1[(1 + m)/2, -p, 1, (3
+ m)/2, -((b*x^2)/a), -((d*x^2)/c)])/(2*c^2*d*(b*c - a*d)*e*(1 + m)*(1 + (b*x^2)/a)^p) - (b*(B*c - A*d)*(1 +
m + 2*p)*(e*x)^(1 + m)*(a + b*x^2)^p*Hypergeometric2F1[(1 + m)/2, -p, (3 + m)/2, -((b*x^2)/a)])/(2*c*d*(b*c -
a*d)*e*(1 + m)*(1 + (b*x^2)/a)^p)

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/
(1 + b*(x^n/a))^FracPart[p]), Int[(c*x)^m*(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 524

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*
((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 525

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPar
t[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p]), Int[(e*x)^m*(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x
] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] &&  !(IntegerQ[
p] || GtQ[a, 0])

Rule 593

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_)*((e_) + (f_.)*(x_)^(n_)), x
_Symbol] :> Simp[(-(b*e - a*f))*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*g*n*(b*c - a*d)*(p +
1))), x] + Dist[1/(a*n*(b*c - a*d)*(p + 1)), Int[(g*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f)
*(m + 1) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c,
d, e, f, g, m, q}, x] && IGtQ[n, 0] && LtQ[p, -1]

Rule 598

Int[(((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n_)))/((c_) + (d_.)*(x_)^(n_)), x_Sy
mbol] :> Int[ExpandIntegrand[(g*x)^m*(a + b*x^n)^p*((e + f*x^n)/(c + d*x^n)), x], x] /; FreeQ[{a, b, c, d, e,
f, g, m, p}, x] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {(e x)^m \left (a+b x^2\right )^p \left (A+B x^2\right )}{\left (c+d x^2\right )^2} \, dx &=\frac {(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^{1+p}}{2 c (b c-a d) e \left (c+d x^2\right )}+\frac {\int \frac {(e x)^m \left (a+b x^2\right )^p \left (2 A b c-a A d (1-m)-a B c (1+m)-b (B c-A d) (1+m+2 p) x^2\right )}{c+d x^2} \, dx}{2 c (b c-a d)}\\ &=\frac {(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^{1+p}}{2 c (b c-a d) e \left (c+d x^2\right )}+\frac {\int \left (-\frac {b (B c-A d) (1+m+2 p) (e x)^m \left (a+b x^2\right )^p}{d}+\frac {(d (2 A b c-a A d (1-m)-a B c (1+m))+b c (B c-A d) (1+m+2 p)) (e x)^m \left (a+b x^2\right )^p}{d \left (c+d x^2\right )}\right ) \, dx}{2 c (b c-a d)}\\ &=\frac {(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^{1+p}}{2 c (b c-a d) e \left (c+d x^2\right )}-\frac {(b (B c-A d) (1+m+2 p)) \int (e x)^m \left (a+b x^2\right )^p \, dx}{2 c d (b c-a d)}-\frac {(a d (A d (1-m)+B c (1+m))-b c (A d (1-m-2 p)+B c (1+m+2 p))) \int \frac {(e x)^m \left (a+b x^2\right )^p}{c+d x^2} \, dx}{2 c d (b c-a d)}\\ &=\frac {(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^{1+p}}{2 c (b c-a d) e \left (c+d x^2\right )}-\frac {\left (b (B c-A d) (1+m+2 p) \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p}\right ) \int (e x)^m \left (1+\frac {b x^2}{a}\right )^p \, dx}{2 c d (b c-a d)}-\frac {\left ((a d (A d (1-m)+B c (1+m))-b c (A d (1-m-2 p)+B c (1+m+2 p))) \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p}\right ) \int \frac {(e x)^m \left (1+\frac {b x^2}{a}\right )^p}{c+d x^2} \, dx}{2 c d (b c-a d)}\\ &=\frac {(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^{1+p}}{2 c (b c-a d) e \left (c+d x^2\right )}-\frac {(a d (A d (1-m)+B c (1+m))-b c (A d (1-m-2 p)+B c (1+m+2 p))) (e x)^{1+m} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} F_1\left (\frac {1+m}{2};-p,1;\frac {3+m}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{2 c^2 d (b c-a d) e (1+m)}-\frac {b (B c-A d) (1+m+2 p) (e x)^{1+m} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\left (\frac {1+m}{2},-p;\frac {3+m}{2};-\frac {b x^2}{a}\right )}{2 c d (b c-a d) e (1+m)}\\ \end {align*}

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Mathematica [A]
time = 0.20, size = 128, normalized size = 0.43 \begin {gather*} \frac {x (e x)^m \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (B c F_1\left (\frac {1+m}{2};-p,1;\frac {3+m}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+(-B c+A d) F_1\left (\frac {1+m}{2};-p,2;\frac {3+m}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )}{c^2 d (1+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((e*x)^m*(a + b*x^2)^p*(A + B*x^2))/(c + d*x^2)^2,x]

[Out]

(x*(e*x)^m*(a + b*x^2)^p*(B*c*AppellF1[(1 + m)/2, -p, 1, (3 + m)/2, -((b*x^2)/a), -((d*x^2)/c)] + (-(B*c) + A*
d)*AppellF1[(1 + m)/2, -p, 2, (3 + m)/2, -((b*x^2)/a), -((d*x^2)/c)]))/(c^2*d*(1 + m)*(1 + (b*x^2)/a)^p)

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {\left (e x \right )^{m} \left (b \,x^{2}+a \right )^{p} \left (B \,x^{2}+A \right )}{\left (d \,x^{2}+c \right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^m*(b*x^2+a)^p*(B*x^2+A)/(d*x^2+c)^2,x)

[Out]

int((e*x)^m*(b*x^2+a)^p*(B*x^2+A)/(d*x^2+c)^2,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*(b*x^2+a)^p*(B*x^2+A)/(d*x^2+c)^2,x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)*(b*x^2 + a)^p*(x*e)^m/(d*x^2 + c)^2, x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*(b*x^2+a)^p*(B*x^2+A)/(d*x^2+c)^2,x, algorithm="fricas")

[Out]

integral((B*x^2 + A)*(b*x^2 + a)^p*(x*e)^m/(d^2*x^4 + 2*c*d*x^2 + c^2), x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**m*(b*x**2+a)**p*(B*x**2+A)/(d*x**2+c)**2,x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*(b*x^2+a)^p*(B*x^2+A)/(d*x^2+c)^2,x, algorithm="giac")

[Out]

integrate((B*x^2 + A)*(b*x^2 + a)^p*(x*e)^m/(d*x^2 + c)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (B\,x^2+A\right )\,{\left (e\,x\right )}^m\,{\left (b\,x^2+a\right )}^p}{{\left (d\,x^2+c\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^2)*(e*x)^m*(a + b*x^2)^p)/(c + d*x^2)^2,x)

[Out]

int(((A + B*x^2)*(e*x)^m*(a + b*x^2)^p)/(c + d*x^2)^2, x)

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